finit element method chapter 1

8/6/2019 finit element method Chapter 1

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Universit de SfaxEcole Nationale

dIngnieurs de SfaxUniversity of Sfax

National Engineering

School of SfaxMthodes des

lments finis

Finite Element

MethodPrpar par Prepared by

Dr. Slim Choura


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Finite Element Method Dr. Slim ChouraPage 1

1Finite element analysis

of one-dimensionalproblems


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The Finite Element Method (FEM) is a technique in which a given spatial domain is represented as a

collection of simple domains, calledfinite elements.

Basic steps

1. Domain discretization.a. Construct the finite element mesh.

b. Number the nodes and elements.

c. Generate the geometric properties (e.g., coordinates and cross-sectional areas)

2. Derivation of element equations.a. Construct the variational formulation of the given differential equation over the typical

element.

b. Assume that a typical dependent variable u is of the form

1

n

i i

i

u u

=

= and substitute it into Step 2a to obtain the element equations in the matrix form

{ } { }e eeK u F = .

c. Derive or select element interpolation functionsi

and compute the element matrices.

3 Assembly of element equations.a. Identify the interelement continuity conditions among the primary variables by relating

element nodes to global nodes.b. Identify the equilibrium conditions among the secondary variables.c. Assemble element equations using Steps 3a and 3b.

4 Imposition of the boundary conditions.a. Identify the specified global primary degrees of freedom.

b. Identify the specified global secondary degrees of freedom (if not already done in Step

3b)

5 Solution of the assembled equations.

6 Post processing.a. Compute the gradient of the solution or other desired quantities from the primary

degrees of freedom computed in Step 5.b. Represent the results in tabular/or graphical form.

Table 1.1 Steps involved in the finite element analysis of a typical problem


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1.1. Model of second-order boundary value problems:

Consider the differential equation describing the axial deformation in a bar shown in Figure 1.1

( ) ( ) ( ) 0 for 0d du

a x c x u q x x Ldx dx

+ = <


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The polynomial approximation of the solution:

( )

1

n

e e e

j j

j

U u x

=

= (1. 3)e

ju are the values of the solution at the nodes and e

j are the approximation functions.

Equation (1.1) in a in a weighted-integral form:

0

B

A

x

x

d duw a cu q dx

dx dx

= + (1. 4)

( )w x is the weight function. Equation (1.4) reduces to

0

BB

AA

x x

xx

dw du dua c wu wq dx wa

dx dx dx

= + (1. 5)

( ) ( )0B

A

x

A A B B

x

dw dua cwu wq dx w x Q w x Q

dx dx

= +

(1. 6)where

A

A

x

duQ adx

=

,

B

B

x

duQ adx

=

(1. 7)

Step 3: Approximations of the Solution:

Linear approximation

( ) ( )

2

1

e e e

j j

j

U x x u

=

=

(1. 8)

where

( )1 1e

e

xx

h = ( )2

e

e

xx

h = (1. 9)

where

A ex x x x x= =


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Quadratic approximation

( ) ( )

3

1

e e e

j j

j

U x x u

=

=

(1. 10)

( ) ( )( )

( )( )

1 2 3

1 1 11 1 ; 1 ; 1

1 1

e e ex x x x x xx x xh h h h h h

= = =

(1. 11)

where 0 1< < and 2 1e e

ex x h= + (i.e.; the second node is chosen arbitrarily at the interior of the

element). For21= (node 2 is at midpoint):

( ) ( ) ( )( )

1 2 31 1 2 4 1 1 21

e e ex x x x x xx x xh h h h h h

= = =

(1. 12)

Note that1

e is equal to 1 at node i and zero at the other two nodes, but varies quadratically

between the nodes.

Step 4: Finite Element Model:

The model can be developed for an arbitrary degree of interpolation:

( )

1

n

e e e

j j

j

u U u x

=

= (1. 13)Following the Rayleigh-Ritz procedure, we substitute (1.13) for u and 1

e , 2e , , en for w into

the weak form (1.6) to obtain n algebraic equations:

( ) ( )1

1 1 1

1 1 1

0

B

A

n n nx eeje e e e e e e e

j j j j j

xj j i

dd

a u c u x q dx x Qdx dx

= = =

= +

( ) ( )2 2 2 21 1 1

0

B

A

n n nx eeje e e e e e e e

j j j j j

xj j i

dda u c u x q dx x Q

dx dx

= = =

= +


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( ) ( )1 1 1

0

B

A

n n nx eeje e e e e e e ei

j i j j i i j j

xj j i


dx dx

= = =

= +

( ) ( )1 1 1

0

B

A

n n nx eeje e e e e e e en

j n j j n n j j

xj j i


dx dx

= = =

= + (1. 14)

The ith algebraic equation can be written as

( )1

1, 2, ... ,

n

e e e eij j i i

j

K u f Q i n

=

= + =

(1. 15)

( )( ) ( )

( ) ( ) ( )0

eh ee

jie e e

ij i j

d xd xK a x c x x x dx

dx dx

= +

, ( ) ( )0eh

e e

i if q x x dx= (1. 16)

or in matrix form

{ } { } { }e e e eK u f Q = + (1. 17)

eK is symmetric (coefficient matrix orstiffness matrix in structural mechanics applications)

{ }ef (source vector orforce vector in structural mechanics applications).

Linear element:

Special case 1: constant eq

{ }

1 1 2 1 1

1 1 1 2 16 2

e ee e e e e

e

a c h q h

K fh

= + =

(1. 18)

Special case 2: ea a x= and ec c= ,

11 1 2 1

1 1 1 22 6

e e e e e e

e

a x x c hK

h

+ +

= +

(1. 19)

Special case 3: a and q are element-wise-constant and 0=c

1 1

2 2

1 1 1

1 1 12

e e

e e e

e e

e

u Qa q h

h u Q

= +

(1. 20)


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Quadratic element:

Special case: a , c and q are element-wise constant

7 8 1 4 2 1

8 16 8 2 16 2303

1 8 7 1 2 4

e e e e

e

a c hK

h

= +

{ }1

46

1

e e eq h

f

=

(1. 21)

Connectivity of Elements:

The assembly of elements is carried out by imposing the following conditions:

1. Continuity of primary variables at connecting nodes:

1

1

e e

nu u+

= (1. 22)

2. Balance of secondary variables at connecting nodes:

1

1

0 0

0 if no external point source is applied

if an external point source of magnitude is applied

e e

nQ QQ Q

+

+ =

(1. 23)

Continuity of the primary variables

1

1 1u U= 1 2

2 1 2u u U= = 2 3

2 1 3u u U= = 1

2 1

E E

Eu u U

= = 2 1E

Eu U += (1. 24)

To obtain 11

e e

nQ Q ++ , we must add the n

th equation of element e to the first equation of element

1e+ , that is, we add

1

n

e e e e

nj j n n

j

K u f Q

=

= + and 1 1 1 11 1 11

n

e e e e

j j

j

K u f Q+ + + +

=

= + to give

eQ1

21

e

21

1e+ e

eQ2 1

1+e

Q 12+

eQ

( )xUe

12

+eu 1

1+eu

21

1+eU

e

21

e+1

2+eU

eU

eu2

eu1

e e+1 e+1 e+2

( )xUe 1+


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( ) ( )1 1 1 11 1 11

n

e e e e e e e e

nj j j j n n

j

K u K u f f Q Q+ + + +

=

+ = + + + 11 0e enf f Q+= + + (1. 25)For a mesh ofElinear elements ( 2=n ), we have

1 1 1

11 12 1 1

1 1 2 2 1 2

21 22 11 12 2 2 1

2 2 3 2 3

21 22 11 3 2 1

1 1

22 11 12 2 1

21 22 1 2

0 0 0

0 0

0 0 0

0 0 0

0 0 0

E E E E E

E

E E E

E

K K U f

K K K K U f f

K K K U f f

K K K U f f

K K U f

+

+ + + +

= + +

1

1

1 2

2 1

2 3

2 1

1

2 1

2

E E

E

Q

Q Q

Q Q

Q Q

Q

+ +

+ +

(1. 26)Recall that the above discussion of assembly is based on the assumption that the elements are

connected in series.

Consider the following structure:

Continuity and force balance conditions:

1 3 2

2 1 2 3u u u U = = = 1 3 2

2 1 2 2Q Q Q P+ + = (1. 27)

To enforce these conditions:

1

1 1u U= 2

1 2u U= 1 3 2

2 1 2 3u u u U = = = 3

2 4u U=

( ) ( ) ( )1 1 1 1 3 3 3 3 2 2 2 2 1 3 2 1 3 221 1 22 2 11 1 12 2 21 1 22 2 2 1 2 2 1 2K u K u K u K u K u K u f f f Q Q Q+ + + + + = + + + + + (1. 28)

h2

h1

P

P

h3

1

2 3

3

3

4

Rigid bar

(constrained to move horizontally)

h2 = h1

2

1

3


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The assembled equations are:

1 1 1 1

111 12 1 1

2 2 2 2

211 12 1 1

1 2 1 3 2 3 1 3 2 1 3 2

321 21 22 11 22 12 2 1 2 2 1 2

3 3 3 3

421 22 2 2

0 0

0 0

0 0

UK K f Q

UK K f QUK K K K K K f f f Q Q Q

UK K f Q

= + + + + + + +

(1. 29)

Step 6: Imposition of Boundary Conditions:

1

1 1 0u U= = 2

1 2 0u U= = 3

2 4 0u U= = (1. 30)

1 3 2

2 1 2 2Q Q Q P+ + = (1. 31)

The displacement 3U and the forces11Q ,

21Q and

32Q are unknown. We obtain

1 1 1

111 12 1

2 2 2

211 12 1

1 2 1 3 2 3

321 21 22 11 22 12

3 3 3

421 22 2

0 0 0

0 0 0

2

0 0 0

UK K Q

UK K Q

PUK K K K K K

UK K Q

=

= = + + =

(1. 32)

Step 5: Solution of Equations:

Condensed equation

( ) ( )1 3 2 1 2 322 11 22 3 21 1 21 2 12 42K K K U P K U K U K U + + = + + (1. 33)

Hence,( )

3 1 3 2

22 11 22

2PU

K K K=

+ +

(1. 34)

The remaining equations can be assembled as:

1 111 11 12 311 12

2 222 21 12 311 12

3 333 32 21 321 22

4

0 0

0 0

0 0

UQ K UK KU

Q K UK KU

Q K UK KU

= =

(1. 35)

Step 6: Post Processing of the Solution:

The solution u as a continuous function of position x is:


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( )

( ) ( )

( ) ( )

( ) ( )

1 1 1

1

2 2 2

1

1

n

j j

j

n

j j

j

n

N N N

j j

j

U x u x

U x u xu x

U x u x

=

=

=

=

=

=

(1. 36)

Depending on the value ofx , the corresponding element equation from (1.66) is used. The derivative

of the solution is obtained by differentiating (1.66):

1

1

1

2

2

1

1

n

j

j

j

n

j

j

j

nN

jNj

j

du

dx

dudu

dxdx

dudx

=

=

=

(1. 37)

1.2. Model of Fourth Order Boundary Value Problem (Bending of Beams) :

1.2.1. The Euler-Bernoulli Beam Element:1.2.1.1. Governing equation:

Transverse deflection w of the beam

L

w

M0

f(x)F0


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( ) ( )2 2

2 2

d d wb x f x

dx dx

=

for 0


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( ) ( )1 1

1

2 2

1 2 1 3 42 20

e e

e ee e

x x

e e e e

e e

x xx x

d v d w dv dvb dx vfdx v x Q Q v x Q Q

dx dxdx dx

+ +

+

+

=

(1. 40)

2

1 2

e

e

x

d wdQ b

dx dx

2

2 2

e

e

x

d wQ b

dx

1

2

3 2

e

e

x

d wdQ b

dx dx+

1

2

4 2

e

e

x

d wQ b

dx+

(1. 41)

Interpolation Functions

In the local coordinate system, the Hermite cubic interpolation functions are

2 3 2 2 3 2

1 2 3 41 3 2 1 3 2e e e e

e e e e e e e

x x x x x x xx x

h h h h h h h

= + = = =

(1. 42)

Finite Element Model

The thi algebraic equation of the finite element model is (for ei

v = )

1 14

22

2 2

1

0

e e

e e

x xeej e e ei

j i i

x xj

ddb dx u fdx Q

dx dx

+ +

=

=

(1. 43)or

4

1

0e e eij j i

j

K u F

=

= (1. 44)where

1 122

2 2

e e

e e

x xeeje e e ei

ij i i i

x x

ddK b dx F fdx Q

dx dx

+ +

= = + (1. 45)The coefficients e

ijK are symmetric (e e

ij jiK K= ). In matrix notation (1.43) can be written as:

11 12 13 14 1 1 1

21 22 23 24 2 2 2

31 32 33 34 3 3 3

41 42 43 44 4 4 4

e e e e e e e

e e e e e e e

e e e e e e e

e e e e e e e

K K K K u f Q

K K K K u f Q

K K K K u f Q

K K K K u f Q

= +

(1. 46)

For the case in which ( )b EI= and f are constant over an element


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{ }

1

2 2

2

3

3

2 24

6 6 63 3

3 2 32

6 6 6123 3

3 3 2

e

e e

e

e ee e e e ee

e

e e e

ee e e e e

h h Q

h h h h h QfhbK F

h h h Q

h h h h h Q

= = +

(1. 47)

Assembly of Element Equations

Continuity conditions:

1 1 1 2

1 1 2 2 3 1 3

1 2 2 2

4 2 4 3 5 4 6

u U u U u u U

u u U u U u U

= = = =

= = = =(1. 48)

Equilibrium conditions:

3 1

4 2

applied external point force

applied external bending moment

e f

e f

Q Q

Q Q

+ =

+ =

(1. 49)

Forces are taken positive acting upward and moments are taken positive acting clockwise.

To impose the equilibrium equations (1.47), we add the third (fourth) equation of the first element to

the first (second) equation of the second element.

1

1 unknownQ = 01 =U

1

2 unknownQ = 2 0U =

3U 1 2

3 10Q Q+ =

4U 1 2

4 20Q Q+ =

2

3 0Q F= 5U

2

4 0Q M= 6U

1

3u

1

4u

1

4Q

1

3Q

1

1u

1

2u

1

2Q

1

1Q

Element 1

23u

2

4u

2

4Q

2

3Q

2

1u

2

2u

2

2Q

2

1Q

Element 2


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[ ]

3

2

1

3

2

1

2

44

2

43

234

233

2

42

2

41

232

231

224

223

214

213

222

144

221

143

212

134

211

133

142

141

132

131

124123

114

113

122121

112

111

++

++=

KK

KK

KK

KK

KK

KK

KKKK

KKKK

KK

KK

KK

KK

KK

KK

K

(1. 50)

{ }

11

12

1 23 1

1 24 2

23

24

F

F

F FF

F F

F

F

+ =

+

(1. 51)

Imposition of Boundary Conditions

Let us consider a two-element cantilever beam

For a mesh of two elements with1 2

1

2h h L h= = = , and constant EI and f

0M

0F 0f

L

z

1 2 3

Global

nodes


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1

1 1

2 2 1

2 2

1 2

3 3 10

3 2 2 2 2 1 24 4 2

2

5 3

2 2

6

6 -3 6 3 0 0 6

3 3 0 02

6 3 6 6 3 3 6 3 6 62

3 3 3 3 122 20 0 6 3 6 3 6

0 0 3 3 2

h h U Q

h h hUh h Q

h h h h U Q Qf hEI

h h h h h hh Uh h h h Q Q

h h U Q

h h hUh h

+ + +

= + + +

2

4Q

(1. 52)

It can be seen that

1 2 1 2

3 1 4 20 0Q Q Q Q+ = + = (1. 53)

2

2

3 02

x L

d wdQ EI F

dx dx=

=

2

2

4 02

x L

d wQ EI M

dx=

=

(1. 54)

1 1

1 1 1 0u w U= = = 1 1

2 1 2 0dw

u Udx

= = = =

(1. 55)

Using (1.99)-(1.101) into (1.98) yields:

1

1 1

12 2

2 2

3 03 2 2 2

4

5 0

2 2

6 0

606 3 6 3 0 0

03 2 3 0 0

12 06 3 12 0 6 32 012 03 0 4 3

60 0 6 3 6 3

0 0 3 3 2

U Qh h

hU Qh h h h

Uh h f hEIh Uh h h h h

U Fh h

hU Mh h h h

=

= = +

(1. 56)

Solution

{ }

2 3

0 0 03

22 40 0 0

2 35

0 0 026

0 0 0

175 3

4

9 6 76

16 12 1212 12 8

h F hM f hU

U hhF M f hU

EIU

h F hM f hUhF M f h

+ +

= =

+ +

(1. 57)

{ }( )

( )

10 0

1 1

1

2 0 0 0

2

2

F f hQF

Q h F f h M

+ = =

+ +

(1. 58)

Post-processing of the solution

The finite element solution as a function of position x is given by


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( )1 1

3 3 4 4

2 2 3 4

3 1 4 2 5 2 6 2

for 0

for 2e

x hU Uw x

h x hU U U U

+=

+ + +(1. 59)

2 3 2 3

1 13 4

2 3 2

2 2

1 2

2 3 3 2

2 2

3 4

3 2

1 3 1 2 1 1 2

3 1 2 1 1 1

x x x xhh h h h

x x x xh

h h h h

x x x xh

h h h h

= =

= =

= + = +

(1. 60)

Exact solution

( ) ( )

( ) ( )

( ) ( )

4 3 2 2

0 0 0 0 0 0

3 2 2

0 0 0 0 0 0

2 2

0 0 0 0 0 0

1 1 1 1

24 6 2 2

1 1 1for 0

6 2 2

1 1

2 2

EIw x f x F f L x M F L f L x

EI x f x F f L x M F L f L x x L

M x f x F f L x M F L f L

= + + + +

= + + + +

= + + + +

(1. 61)

10 mkN24 -f = , kN060 =F , m3=L ,

mkN00 =M ,26 mkN10002 -E = , 46 mm1092 =I

As expected, the finite element solution for w and coincides with the exact solution at the nodes.

At other points, the difference is less than 2%.

w (m) - M/EI(kN m)

x (m) FEM Exact FEM Exact FEM Exact

0.00 0.00 0.00 0.00 0.00 0.0489 0.0497

0.1875 0.0008 0.0008 0.0088 0.0089 0.0452 0.04550.3750 0.0033 0.0033 0.0169 0.0171 0.0415 0.0414

0.5625 0.0071 0.0072 0.0244 0.0245 0.0378 0.03750.7500 0.0124 0.0124 0.0311 0.0311 0.0341 0.0338

0.9375 0.0188 0.0188 0.0372 0.0371 0.0305 0.0301

1.1250 0.0263 0.0263 0.0426 0.0425 0.0268 0.02661.3125 0.0347 0.0347 0.0472 0.0471 0.0231 0.0234

1.5000 0.0439 0.0439 0.0512 0.0512 0.0194 0.0202

1.6875 0.0539 0.0539 0.0546 0.0547 0.0169 0.01711.8750 0.0644 0.0644 0.0575 0.0576 0.0144 0.0143

2.0625 0.0754 0.0755 0.0600 0.0600 0.0118 0.0115

2.2500 0.0868 0.0869 0.0620 0.0620 0.0093 0.0089

2.4375 0.0986 0.0987 0.0635 0.0634 0.0068 0.0065

2.6250 0.1106 0.1107 0.0645 0.0644 0.0043 0.0042

2.8025 0.1228 0.1228 0.0651 0.0650 0.0017 0.0020

3.0000 0.1350 0.1350 0.0652 0.0652 -0.0008 0.0000


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1.2.2. Plane Truss and Euler-Bernoulli Frame Elements:Structures composed of bar elements and beam elements are classified as truss and frame structures.

Truss structure:

Frame structure:

Element degrees of freedom Global degrees of freedom

x

z

VF

HF

x

oF

A pin connection

Members rotatez

F

0f

x

z

z

x

2u

5u

1u

4u

3u

6u

x

z

z

x

2u

5u

1u

4u

3u

6u

x

z


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{ } { }e e eK u F = (1. 62)

{ }

1

1

1

2

2

2

e

u

w

uu

w

=

, { }

1

2

2

3

4

5

2

6

1/ 2

1/ 2

1/12

1/ 2

1/ 2

1/12

e

e

e

e

e

e

e

qh Q

fh Q

fh QF

qh Q

fh Q

fh Q

= +

(1. 63)

and 0qq = and 0ff = are constants over an element:

2 2

3

2 2

0 0 0 0

0 6 3 0 6 3

0 3 2 0 32

0 0 0 0

0 6 3 0 6 3

0 3 0 3 2

e

h h

h h h hEIK

h

h h

h h h h

=

,2

2

Ah

I = (1. 64)

Let u and w denote the axial and transverse displacements referred to the local coordinate system

( )zyx ,, :

cos 0 sin

0 1 0

sin 0 cos

x x

y y

z z

=

(1. 65)

where the angle is measured counter-clockwise from the x axis to the x axis. In addition

cos sin 0sin cos 0

0 0 1

u uw w

=

(1. 66)

Therefore,


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1 1

2 2

3 3

4 4

5 5

6 6

cos sin 0 0 0 0

sin cos 0 0 0 0

0 0 1 0 0 0

0 0 0 cos sin 0

0 0 0 sin cos 0

0 0 0 0 0 1

e e

u u

u u

u u

u u

u u

u u

=

(1. 67)

or

{ } { }e e eu T u = (1. 68)

Analogously

{ } { }e e eF T F = (1. 69)

Using (1.67) and (1.68) in the element equations

{ } { }e e eK u F = (1. 70)

we obtain

{ } { }T

e e e e eT K T u F = (1. 71)

which gives the element stiffness matrix

{ } { }=T T

e e e e e e eK T K T F T F = (1. 72)

It can be shown that the above matrix has the following form:

( )

( ) ( )

( ) ( ) ( )

2 2

2 2

2

2 2 2 23

2 2 2

cos 6sin

6 cos sin sin 6cos symmetric

3 sin 3 cos 22

cos 6sin 6 cos sin 3 sin cos 6sin

6 cos sin sin 6cos 3 cos 6 cos sin sin

e

h h hEI

Khh

h

+

+

= + +

+ 2

2 2

6cos

3 sin 3 cos 3 sin 3 cos 2

e

h h h h h h

+

(1. 73)

{ }

1 2

1 2

2

3

4 5

4 5

6

cos sin

sin cos

2 cos sin

sin cos

e

ee e

e

e

F F

F F

FA hF

I F F

F F

F

+

= =

+

(1. 74)


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Example 1:

Consider the three-member truss shown in Figure 1. 15(a). All members of the truss have identical

cross-sectional areaA and modulusE. We wish to determine the horizontal and vertical

displacements at the joint 3 and the forces in each member of the structure.

(a)

(b)

Figure 1.15. (a) Geometry and loading (b) element numbers, global node numbers of element nodes

and element nodal forces in the element coordinates

2P

P

x

z

2

1

3

L

L 3

1

2

45o

90o

6U

5U

2U

1U

4U

3U 4

3

2

1

3

4

2

1

2

1

3

4

1

23

2

1

1

3

32


22/25


Since all joints are hinged, and the applied forces are acting at the nodes, the members are subjected

to only axial forces. Hence, the structure is a truss. We use three finite elements to model the

structure. Any further subdivision of the members does not add to the accuracy, because for all truss

problems the finite element solutions are exact. This is a consequence of the fact that all truss

members with constant cross-section are governed by the homogeneous differential equation

2

20

d uEA

dx= (1. 75)

whose solution is of the form ( ) 1 2u x c x c= + . Thus, linear interpolation of the displacements should

give the exact result.

There are two degrees of freedom, horizontal and vertical displacements, at each node of the

element. The element stiffness matrix in the local coordinate system is:

1 0 1 0

0 0 0 0

1 0 1 0

0 0 0 0

e e e

e

A EK

h

=

(1. 76)

The transformed stiffness matrix of the elemente

in the global coordinate system is given by:

2 2

2 2

2 2

2 2 2

cos cos sin cos cos sin

cos sin sin cos sin sin

cos cos sin cos cos sin

cos sin sin cos sin sin

e e e

e

E AK

h

=

(1. 77)

which is obtained from (1.127) by deleting the rows and columns corresponding to the bending

degrees of freedom and setting all bending stiffnesses to zero.

The element data for the problem are as follows:

Elementnumber

Global nodesof the element

Geometricproperties

Materialproperty

Orientation

1 2 3 A, Lh =1 EO0=

2 1 3 A, Lh 22 = EO45=

3 1 2 A, Lh =1 EO90=

The assembled stiffness and force coefficients are given by:


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[ ]

2 3 2 3 3 3 2 2

11 11 12 12 13 14 13 14

2 3 3 3 2 2

22 22 23 24 23 24

3 1 3 1 1 1

33 11 34 12 13 143 1 1 1

44 22 23 24

2 1 2 1

33 33 34 34

2 1

4 44

symmetric

K K K K K K K K

K K K K K K

K K K K K K K

K K K K

K K K K

K K

+ +

+ + +

= + + +

+

(1. 78)

The force vector can be written:

{ }

3 2

1 1

3 2

2 2

3 1

3 1

3 1

4 21 2

3 3

1 2

4 4

Q Q

Q Q

Q QF

Q Q

Q Q

Q Q

+

+ +

= +

+

+

(1. 79)

Substituting the element data into (1.132), we get:

[ ]

0.3536 0.3536 0 0 0.3536 0.3536

1.3536 0 1 0.3536 0.3536

1 0 1 0

1 0 0

symmetric 1.3536 0.3536

0.3536

K

=

(1. 80)

The specified displacement degrees of freedom are

1 2 3 4 0U U U U = = = = (1. 81)

The condensed equations for the unknown displacements and forces are

5

6

1.3536 0.3536 2

0.3536 0.3536

PUEA

PL U

=

(1. 82)

3 2

1 1

3 2

52 2

3 1

63 1

3 1

4 2

0.3536 0.3536

0.3536 0.3536

1 0

0 0

Q Q

UQ Q EA

L UQ Q

Q Q

+ +

= +

+

(1. 83)

where, for example, 3 21 1Q Q+ is the horizontal force and

3 2

2 2Q Q+ the vertical force at node 1.

Solving (1.136) and (1.137) yields:

P2

P


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5

3PLU

EA= ( )6 3 2 2 5.828

PL PLU

EA EA= + =

2 3

1 1 1 F Q Q P= + = 2 F P= 3 3F P= 4 0F = (1. 84)

Example 2:

The frame structure shown in Figure 1.16 is to be analyzed for displacements and forces.

(a) (b)

Figure 1.16. (a) geometry and loading (b) finite element discretization

The geometric and material properties of each element are as follows:

Element 1:

L = 144 in, A = 10 in2, I= 10 in

4, 0cos = , 1sin = ,E= 10

6psi, Pf

7211 = lb in

-1.

Element 2:

L = 180 in, A = 10 in2, I= 10 in4, 0.8cos = , 0.6sin = ,E= 106 psi, 02 =f .

The assembled stiffness matrix and force vectors are obtained by superimposing the last three rows

and columns of element 1 on the first three rows and columns of element 2; i.e., the 33 submatrix

associated with rows and columns 4, 5 and 6 of element 1, and 33 submatrix associated with rows

and columns 1, 2 and 3 of element 2 overlap in the global stiffness matrix.

The known geometric boundary conditions are:

21Q

16Q

4P

)ft(lb

6

1 1P

2P

x

y

1

2

3

6 ft

12 ft

9 ft

12 ft1

2

3

1 3

14Q

15Q

11Q

12Q 1

3Q

2

4

2

23Q

22Q

25Q

24Q

2

6

Q


25/25


01 =U , 02 =U , 03 =U , 07 =U , 08 =U , 09 =U (1. 85)

The force boundary conditions are

1 24 1 0Q Q+ = , 1 25 2 2Q Q P+ = , 1 26 3 0Q Q+ = (1. 86)

Since all specified values of the known boundary conditions on the primary variables are zero, the

condensed equations for the unknown global displacement degrees of freedom are:

4

5

5

6

0.3560 0.2666 0.0178 1

10 0.2666 0.8846 0.0148 4

0.0178 0.0148 5.0000 48

U

U P

U

=

(1. 87)

whose solution is

( )in100.839 44 PU = ( )in100.681 45 PU = ( )rad100.961 44 PU = (1. 88)

The reactions and forces in each member in the global coordinates can be computed from the

element equations

{ } { } { }e e e eQ K u f = (1. 89)

The forces eQ can be transformed to those in the element coordinate system by means of

{ } { }e e eQ T Q = (1. 90)

we obtain

{ }1

4.731

0.725

10.900

4.731

1.275

50.450

Q P

=

, { }2

1.458

0.180

21.550

1.458

0.180

10.870

Q P

=

(1. 91)

finit element method chapter 1

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